The idea of gravity has a long and interesting history intertwined with our evolving ideas about the solar system. It was Newton who—along with discovering his three laws of motion—first recognized that gravity is an attractive, long-range force between any two objects.

FIGURE 6.6 shows two objects with masses $m_1$ and $m_2$ separated by distance $r$. Each object pulls on the other with a force given by Newton’s law of gravity:

where $G=6.67\times {10}^{-11}{\text{N m}}^2\text{/}{\text{kg}}^2$, called the gravitational constant, is one of the basic constants of nature. Notice that gravity is not a constant force—the force gets weaker as the distance between the objects increases.

FIGURE 6.6 Newton’s law of gravity.

The gravitational force between two human-sized objects is minuscule, completely insignificant in comparison with other forces. That’s why you’re not aware of being tugged toward everything around you. Only when one or both objects are planet-sized or larger does gravity become an important force. Indeed, Chapter 13 will explore in detail the application of Newton’s law of gravity to the orbits of satellites and planets.

For objects moving near the surface of the earth (or other planet), things like balls and cars and planes that we’ll be studying in the next few chapters, we can make the flat-earth approximation shown in FIGURE 6.7. That is, if the object’s height above the surface is very small in comparison with the size of the planet, then the curvature of the surface is not noticeable and there’s virtually no difference between r and the planet’s radius $R$. Consequently, a very good approximation for the gravitational force of the planet on mass $m$ is simply

FIGURE 6.7 Gravity near the surface of a planet.

The magnitude or size of the gravitational force is $F_{\text{G}}=mg$, where the quantity g—a property of the planet—is defined to be

Also, the direction of the gravitational force defines what we mean by “straight down.”

But why did we choose to call it g, a symbol we’ve already used for free-fall acceleration? To see the connection, recall that free fall is motion under the influence of gravity only. FIGURE 6.8 shows the free-body diagram of an object in free fall near the surface of a planet. With $\vec{F}{}_{\text{net}}=\vec{F}{}_{\text{G}}$, Newton’s second law predicts the acceleration to be

Because $g$ is a property of the planet, independent of the object, all objects on the same planet, regardless of mass, have the same free-fall acceleration. We introduced this idea in Chapter 2 as an experimental discovery of Galileo, but now we see that the mass independence of $\vec{a}{}_{\text{free fall}}$ is a prediction of Newton’s law of gravity.

FIGURE 6.8 The free-body diagram of an object in free fall.

But does Newton’s law predict the correct value, which we know from experiment to be $g=\left|a_{\text{free fall}}\right|=9.80\text{m}\text{/}{\text{s}}^2?$ We can use the average radius $(R_{\text{earth}}=6.37\times {10}^6\text{m})$ and mass $(M_{\text{earth}}=5.97\times {10}^{24}\text{kg})$ of the earth to calculate

You should convince yourself that $\text{N}\text{/}\text{kg}$ is equivalent to $\text{m}\text{/}{\text{s}}^2$, so $g_{\text{earth}}=9.82\text{m}\text{/}{\text{s}}^2$.

Newton’s prediction is very close, but it’s not quite right. The free-fall acceleration would be $9.82\text{m}\text{/}{\text{s}}^2$ on a stationary earth, but, in reality, the earth is rotating on its axis. The “missing” $0.02\text{m}\text{/}{\text{s}}^2$ is due to the earth’s rotation, a claim we’ll justify when we study circular motion in Chapter 8. Because we’re on the outside of a rotating sphere, rather like being on the outside edge of a merry-go-round, the effect of rotation is to “weaken” gravity.

Strictly speaking, Newton’s laws of motion are not valid in an earth-based reference frame because it is rotating and thus is not an inertial reference frame. Fortunately, we can use Newton’s laws to analyze motion near the earth’s surface, and we can use $F_{\text{G}}=mg$ for the gravitational force if we use $g=\left|a_{\text{free fall}}\right|=9.80\text{m}\text{/}{\text{s}}^2$ rather than $g=g_{\text{earth}}$. (This assertion is proved in more advanced classes.) In our rotating reference frame, $\vec{F}{}_{\text{G}}$ is the effective gravitational force, the true gravitational force given by Newton’s law of gravity plus a small correction due to our rotation. This is the force to show on free-body diagrams and use in calculations.